Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / Paper 1 / 18

Let $I_n$ denote the integral with the cosine factor and $J_n$ the one with the sine factor. In integration by parts, take $u=\cos (nx)$ and $dv=e^{x}dx$. ThenThe integer restriction gives $\cos (n\pi )=(-1{)}^n$.

For $J_n$, take $u=\sin (nx)$ and $dv=e^{x}dx$ in integration by parts. This yieldssince $\sin (n\pi )=\sin 0=0$ for every integer $n$.

Put $D_n=(-1{)}^n{e}^{\pi }-1$. The two linear equations from parts (i) and (ii) give $I_n=D_n+n{J}_n$ and $J_n=-n{I}_n$, so $(1+n^2)I_n=D_n$. ThereforeThese formulas include $n=0$, when $I_0=e^{\pi }-1$ and $J_0=0$.

A Fourier series represents a periodic function as a constant plus sine and cosine waves whose frequencies are integer multiples of its fundamental frequency. For period $2L$, use the conventionwhere the Fourier coefficients areThe coefficient formulas follow by multiplying by a basis sine or cosine and using their orthogonality over a full period. A Fourier series need not equal an arbitrary function at every point. The pointwise convergence of a piecewise smooth Fourier series theorem says that for a piecewise continuously differentiable function, the Fourier series converges at each point to the average of the two one-sided limits, and hence to $f(x)$ wherever $f$ is continuous.

The function is even, so each product with a sine is an odd function and all the Fourier coefficients $b_n$ vanish. Its constant coefficient isFor $n\ge 1$, part (a) gives the remaining Fourier coefficients:Consequently the Fourier cosine series isThe periodic extension is continuous, including at $\pm \pi$, and piecewise continuously differentiable. Thus the Fourier series equals $f$ everywhere. Its coefficients are also bounded in magnitude by a constant times $1/n^2$, which gives absolute convergence.

At $x=\pi /2$, all odd-indexed cosine terms vanish, while $\cos (2k\pi /2)=(-1{)}^k$. Thus the Fourier series from part (c) givesDivide by $2(e^{\pi }-1)/\pi$ and rearrange. The hyperbolic sine identity $e^{\pi }-1=2e^{\pi /2}\sinh (\pi /2)$ yieldsThe evaluation is legitimate by the convergence established in part (c).